377 + Interview Questions in STRAIGHT-LINES IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Consider the lines `L_(1) -=3x-4y+2=0 " and " L_(2)-=3y-4x-5=0.` Now, choose the correct statement(s).A. The line x+y=0 bisects the acute angle between `L_(1) "and " L_(2)` containing the origin.B. The line x-y+1=0 bisects the obtuse angle between `L_(1) " and " L_(2)` not containing the origin.C. The line x+y+3=0 bisects the obtuse angle between `L_(1) " and " L_(2)` containing the origin.D. The line x-y+1=0 bisects the acute angle between `L_(1) " and " L_(2)` not containing the origin.
Answer» Correct Answer - A::B We have `L_(1) -= 3x-4y+2=0` `L_(2)-=4x-3y+5 =0` Here, `a_(1)a_(2) + b_(1)b_(2) = (3)(4) + (-4)(-3) = 24 gt 0` So, acute angle bisector is 3x-4y+2=4x-3y+5 or x+y+3=0 This bisector goes through the regions where expressions 3x-4y+2 and 4x-3y+5 have the same sign. Also, this is the bisector of anlge which contains origin.

Discussion

102.	Find angles between the lines `sqrt(3)x+y=1`and `x+sqrt(3)y=1`.
Answer» `sqrt3x+y=1 Rightarrow y=-sqrt3x+1` and `x+sqrt3y=1 Rightarrow y=-(1)/(sqrt3)x+(1)/(sqrt3)` `therefore m_(1)=-sqrt3 and m_(2)=(-1)/(sqrt3)` Let `theta` the angle between the given lines. Then, `tan theta=\|(m_(2)-m_(1))/(1+m_(1)m_(2))\|=\|((-1)/(sqrt3)+sqrt3)/({1+(-sqrt3)xx((-1)/(sqrt3))})\|=\|(2)/(sqrt3)xx(1)/(2)\|=\|(1)/(sqrt3)\|=(1)/(sqrt3)` `Rightarrow theta=30^(@) and (180^(@)-theta)=(180^(@)-30^(@))=150^(@)` Hence, the angles between the given lines are `30^(@) and 150^(@)`

Discussion

103.	Point R (h, k) divides a line segment between the axes m the ratio `1: 2`. Find equation of the line.
Answer» `x/a + y/b = 1` given point R Is`R(h,k) = [(20 + 1a)/(1+2) , (2b+ 10)/(1+2)]` `h=a/3 & k=(2b)/3` `a=3h & b= (3k)/2` equation of line will be `x/(3h) + y/((3k)/2) = 1` `x/h + (2y)/k = 3` answer

Discussion

104.	If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Answer» Given, x-intercept of line L is 2 and y-intercept of line L is 3 ∴ The line L intersects X-axis at (2, 0) and Y axis at (0,3). ∴ The line L passes through (2, 0) and (0, 3). Slope of line L = \(\frac {y_2-y_1}{x_2-x_1} = \frac {3-0}{0-2}= \frac{-3}{2}\)

104.

If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.

Answer»

Given, x-intercept of line L is 2 and y-intercept of line L is 3

∴ The line L intersects X-axis at (2, 0) and Y axis at (0,3).

∴ The line L passes through (2, 0) and (0, 3).

Slope of line L = \(\frac {y_2-y_1}{x_2-x_1} = \frac {3-0}{0-2}= \frac{-3}{2}\)

Discussion

105.	Find the distance of the point (2, 3) from the line y=4.
Answer» Correct Answer - 1 unit

Discussion

106.	Reduce the equation `x+y-sqrt2=0` to the normal form `x cos alpha+y sin alpha=p`, and hence find the values of `alpha and p`.
Answer» Correct Answer - `xcos45^(@)+ysin45^(@)=1;alpha=45^(@),p=1`

Discussion

107.	Reduce the equation `sqrt3+y+2=0` to the normal form `x cos alpha+y sin alpha=p`, and hence find the value of `alpha and p`.
Answer» We have `sqrt3x+y+2=0 Rightarrow -sqrt3x-y=2 " "["keeping constant +ve"]` `Rightarrow ((-sqrt3)/(2))x+((-1)/(2))y=1 ` `["on dividing throughout by"sqrt((sqrt3)^(2)+1^(2))]` `Rightarrow x cos alpha+y sin alpha=p,"where "cos alpha=(-1)/(2), sin alpha=(-sqrt3)/(2) and p=(5)/(2)` Since, `cos alpha lt 0 and sin alpha lt 0, alpha lies in third quadrant. Now, `tan alpha=(sin alpha)/(cos alpha)=((-sqrt3)/(2))xx(-2)=sqrt3=tan(180^(@)+60^(@))=tan 240^(@)` Thus, ``alpha=240^(@) and p=(5)/(2)` Hence, the given equation in normal form is `x cos 240^(@)+y sin 240^(@)=(5)/(2)`

Discussion

108.	Reduce each of the followringequations to normal form: (i)x+y-2=0 (ii) `x+y+sqrt2=0` (iii)x+5=0 (iv)2y-3=0 (v)4x+3y-9=0
Answer» Correct Answer - `(i)x cos 45^(@)+ysin45^(@)=sqrt2` `x cos 225^(@)+ysin225^(@)=1` `(iii)x cos180^(@)+y sin 180^(@)=5 ` `(v)x cos alpha+y sin alpha=p, "where "cos alpha=(4)/(5), sin alpha=(3)/(5) and p=(9)/(5)`

Discussion

109.	Find the distance of the point (-4,3) from the line 4(x+5)=3(y-6)
Answer» Correct Answer - `(13)/(5)"units"`

Discussion

110.	The coordinates of the image of the origin O with respect to the line `x+y+1=0` areA. (-1/2,-1/2)B. (-2,-2)C. (1,1)D. (-1,-1)
Answer» Correct Answer - D

Discussion

111.	Reduce the equation y+4=0 to the normal form `x cos alhpha+y sin alpha=p` and hence find the values of `alpha and p`.
Answer» We have `y+4=0 Rightarrow -y=4" "["keeping constant positive"]` `Rightarrow x cos alpha+y sin alpha=p,"where cos "alpha=0,sin alpha=-1 and p=4` Now, `(cos alpha=0 and sin alpha=-1) Rightarrow alpha=270^(@)` Hence, the required normal form of hte given equation is `x cos 270^(@)+y sin 270^(@)=4` `"Clearly", alpha=270^(@) and p=4`

Discussion

112.	If the points `(1, 2)` and `(3, 4)` were to be on the opposite side of the `3x - 5y + alpha = 0`, then:A. `7 lt a lt 11`B. a=7C. a=1D. `a lt 7 " or " a gt 11`
Answer» Correct Answer - D

Discussion

113.	Find the distance of the point (-2,3) from the line 12x=5y+13.
Answer» Correct Answer - 4 units

Discussion

114.	Find the distance of the point (4,1) from the line `3x-4y+12=0`
Answer» Clerly, the required distance is the length of perpendicular from the point P(4,1) on the line 3x-4y+12=0 Let the required distance of d. Then, `d=(\|3xx4-4xx1+12\|)/(sqrt(3^(2)+(-4)^(2)))=(20)/(5)"units=4units"`

Discussion

115.	Find the distance of the point (-1,1) and the given line is `12x-5y+82=0`
Answer» The given point is P(-1,1) and the given line is 12x-5y+82=0. Let the required distance be d. The, d=length of the perpendicular from P(-1,1) on the line 12-5y+82=0 `=(\|12xx(-1)-5xx1+82\|)/(sqrt((12)^(2)+(-5)^(2)))=(65)/(sqrt169)=(65)/(13)=5units`

Discussion

116.	Find the length of the perpendicular from the point (a,b) to the line `(x)/(a)+(y)/(b)=1`
Answer» The given point is P(a,b) and the given line is `bx+ay-ab=0` Let d be the length of the perpendicular from P(a,b) to the line bx+ay-ab=0 `"Then",d=(\|bxxa+axxb-ab\|)/(sqrt(b^(2)+a^(2)))=(\|ab\|)/(sqrt(a^(2)+b^(2)))"units"`

Discussion

117.	Find the length of the perpendicular from the origin to the line 4x+3y-2=0
Answer» The given point is P(0,0) and the given line is 4x+3y-2=0 Let d be the length of the perpendicular from P(0,0) to the line 4x+3y-2=0 "Then"`"Then",d=(\|4xx0+3xx0-2\|)/(sqrt(4^(2)+3^(2)))=(2)/(5)"units"`

Discussion

118.	If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that `1/(p^2)=1/(a^2)+1/(b^2)`.
Answer» The equation of the line making intercepts a and b on the axes is given by `(x)/(a)+(y)/(b)=1, i.e. (x)/(a)+(y)/(b)=1=0.....(i)` Since p is the length of the perpendicular from O(0,0) to line (i), we have `p=(\|(1)/(a)xx0+(1)/(b)xx0-1\|)/(sqrt((1)/(a^(2))+(1)/(b^(2))))=(\|-1\|)/(sqrt((1)/(a^(2))+(1)/(b^(2))))=-(1)/(sqrt((1)/(a^(2))+(1)/(b^(2))))` `Rightarrow p^(2)=(1)/(((1)/(a^(2))+(1)/(b^(2))))=(a^(2)b^(2))/((b^(2)+a^(2)))` `Rightarrow (1)/(p^(2))=((b^(2)+a^(2)))/(a^(2)b^(2))=((b^(2))/(a^(2)b^(2))+(a^(2))/(a^(2)b^(2)))=((1)/(a^(2))+(1)/(b^(2)))` `"Hence"(1)/(p^(2))=(1)/(a^(2))+(1)/(b^(2))`

Discussion

119.	Find the coordinates of the midpoint of the ilne segment joining the points, `A(-2,-5) and B(3,-1)`
Answer» Let the required point be P(x,y). Then, `x=((-2+3))/(2)=(1)/(2),y=((-5)+(-1))/(2)=-3` Hence, the required point is `((1)/(2),-3)`

Discussion

120.	Find what the following equation become when theorigin is shifted to the point (1,1): `x y-x-y+1=0`
Answer» Correct Answer - xy=0

Discussion

121.	Find what the following equation become when theorigin is shifted to the point (1,1): `x^2-y^2-2x+2y=0`
Answer» Correct Answer - `x^(2)-y^(2)=0`

Discussion

122.	Find what the following equation become when theorigin is shifted to the point (1,1): `x y-y^2-x+y=0`
Answer» Correct Answer - `y^(2)-xy=0`

Discussion

123.	The equation `x^2+xy-3x-y+2=0` beome when the origin is shifted to the point `(1,1)` is
Answer» Correct Answer - `x^(2)+xy=0`

Discussion

124.	A line is drawn through the point (1, 2) to meet the coordinate axesat P and Q such that it forms a triangle OPQ, where O is the origin. If thearea of the triangle OPQ is least, then the slope of the line PQ is(1) `-1/4`(2) `-4`(3) `-2`(4) `-1/2`
Answer» Line PQ `(y-2)=m(x-1)` If x=0 then y=2-m if y=0 then x=`1-2/m` Area of `/_OPQ=1/2(2-m)((m-2)/m)` after differentiate with respect to m `(deltaA)/(deltam)=delta/(deltam)(-1/2(m-2)^2/m)` putting`(deltaA)/(deltam)=0` and solving it we get `m^2=4` `m=pm2`

Discussion

125.	If the point `A` is symmetric to the point `B(4,-1)` with respect to the bisector of the first quadrant then `AB` isA. `3sqrt(2)`B. `5sqrt(2)`C. `7sqrt(2)`D. `9sqrt(2)`
Answer» Correct Answer - B It is given that the point Q is symmetric to P(4,-1) with respect to y = x. So, coordinates of Q are (-1,4). `therefore " " PQ = sqrt((4-(-1))^(2)+(-1-4)^(2))= 5sqrt(2)`

Discussion

126.	Let the algebraic sum of the perpendicular distances from the points `(2,0),(0,2)a n d(1,1)`to a variable straight line be zero. Then the line pass through a fixedpoint whose coordinates are`(1,1)`b. `(2,2)`c. `(3,3)`d. `(4,4)`A. `(-1,1)`B. `(1,1)`C. `(1,-1)`D. `(-1,-1)`
Answer» Correct Answer - B Let the equation of the line be `ax + by + c =0` Then, according to the given condition, we have `(2a + 0b + c)/(sqrt(a^(2) +b^(2)))+(0+2b +c)/(sqrt(a^(2)+b))+(a+b+c)/(sqrt(a^(2)+b^(2))) =0` `rArr 3a + 3b + 3c =0 rArr a+b + c=0` This shows that `ax + by + c=0` passes through the fixed point (1,1).

Discussion

127.	Find the equations of the lines, which cut-off intercepts on the axeswhose sum and product are `1`and `-6`, respectively.
Answer» Let the required equation be `(x)/(a)+(y)/(b)=1`. Then x-intercept=a and y-intercept-b `therefore a+b…..(i)` and ab=-……..(ii) Putting `b=(-6)/(a)` from (ii) in (i), we get `a-(6)/(a)=1 Leftrightarrow a^(2)-a-6=0 Leftrightarrow (a-3)(a+2)=0` `Leftrightarrow a=3 or a=-2` `"Now", a=3 Leftrightarrow b=(1-a)=(1-3)=-2` `"And", a=-2 Leftrightarrow b=(1-a)=(1+2)=3` `therefore` the required equation is `(x)/(3)+(y)/(-2)=1 or (x)/(-2)+(y)/(3)=1` `i.e. 2x-3y-6=0 or 3x-2y+6=0`

Discussion

128.	`A(1,3)and c(-2/5,-2/5)`are the vertices of a `DeltaABCand`the equation of the angle bisector of `/_ABC` is `x+y=2.`A. (3/10, 17/10)B. (17/10, 3/10)C. (-5/2, 9/2)D. (1,1)
Answer» Correct Answer - C Vertex B is the point of intersection of 7x+3y+4=0 and x+y-2, i.e., `B-=(-5//2, 9//2)`.

Discussion

129.	The point `(4,1)` undergoes the following two successive transformations (i) Reflection about the line `y=x` (ii) Translation through a distance 2 units along the positive X-axis. Then the final coordinate of the point areA. `(4,3)`B. `(3,4)`C. `(1,4)`D. `((7)/(2),(7)/(2))`
Answer» Correct Answer - B Let the reflection of `A(4,1)` in y=x is B (h,k). Now, mid -point of AB is `((4+h)/(2),(1+k)/(2))` which lies on y=x. i.e. `(4+h)/(2)=(1+k)/(2)rArrh-k=-3` .......(i) So, the slope of liney=xis 1. `therefore` Slope of `AB=(h-1)/(k-1)` `rArr 1.((h-4)/(k-4))=-1` `rArr h-4=1-k` `rArrh+k=5` and `h-k=-3` `2h=2rArrh=1` On putting h=1in Eq.(ii) we get `k=4` So, the point is `(1,4)` Hence, after translation the point is `(1+2,4)` or `(3,4)`

Discussion

130.	Find the equation of the line passing through the point P(-3,5) and perpendicular to the line passing through the points A(2, 5) and B(-3, 6).
Answer» Correct Answer - 5x-y+20=0 Slope of AB=`m_(1)=(6-5)/(-3-2)=(-1)/(5)` Now, `m_(1)m_(2)=-1 Leftrightarrow (-1)/(5)xxm_(2)=-1 Leftrightarrow m_(2)=5` `therefore` required equation is `(y-5)/(x+3)=5 Leftrightarrow 5x-y+20=0`.

Discussion

131.	If a, c, b are in G.P then the line `ax + by + c= 0`A. has a fixed directionB. always passes through a fixed pointC. forms a triangle with the axes whose area is constantD. always cuts intercepts on the axes such that their sum is zero
Answer» Correct Answer - C

Discussion

132.	Straight lines `(x)/(a)+(y)/(b)=1, (x)/(b)+(y)/(a)=1,(x)/(a)+(y)/(b)=2 and (x)/(b)+(y)/(a)=2` form a rhombus of area ( in square units)A. `(ab)/(\|a^(2)-b^(2)\|)`B. `(ab)/(a^(2)+b^(2))`C. `(a^(2)b^(2))/(a^(2)+b^(2))`D. `(a^(2)b^(2))/(\|a^(2)-b^(2)\|)`
Answer» Correct Answer - D The equations of the four sides are `(x)/(a)+(y)/(b)=1" "…(i)" "(x)/(b)+(y)/(a)=1` `(x)/(a)+(y)/(b)=2" "…(iii) " "(x)/(b)+(y)/(a)=2" "…(iv)` Clearly , (i) , (iii) and (ii), (iv) form two sets of parallel lines. So, the four lines form a parallelogram. Area of rhombus `=\|((2-1)(2-1))/({:((1)/(a),(1)/(b)),((1)/(b),(1)/(a)):})\|=\|(1)/((1)/(a^(2))-(1)/(b^(2)))\|` `implies` Area of the rhombus `=(a^(2)b^(2))/(\|b^(2)-a^(2)\|)`

Discussion

133.	Equation of the line passing through `(1,2)` and parallel to the line `y=3x-1` isA. `y+2=x+1`B. `y+2=3(x+1)`C. `y-2=3(x-1)`D. `y-2=x-1`
Answer» Correct Answer - C Since, the line passes through `(1,2)` and parallel to the line `y=3x-1`. So, slope of the required line `m=3` [`because` of `y=3x-1` is3] Hence, the equation of line is `y-2=3(x-1)`

Discussion

134.	Find the image of the point `(4,-13)`in the line `5x+y+6=0.`A. (1,-14)B. (6,-15)C. (-1,-14)D. none of these
Answer» Correct Answer - C Let `Q(x_(1),y_(1))` be the image of the point P(4,-13) in the line mirror `5x+y+6=0` . Then, the coordinates of Q are given by `(x_(1)-4)/(5)=(y_(1)+13)/(1)=(-2(5xx4-13+6))/(5^(2)+1^(2))` `implies (x_(1)-4)/(5)=(y_(1)-13)/(1)=1` `implies x_(1)-4=-5 and y_(1)+13=-1` `implies x_(1)=-1 and y_(1)=-14` Hence, the image of the point P(4,-13) in the line mirror `5x+y+6=0` is (-1,-14).

Discussion

135.	For specifying a straight line, how many geomatrical parameters should be known ?A. 1B. 2C. 4D. 3
Answer» Correct Answer - B Equation of straight lines are `y=mx+ c`, Parameter =2 .....(i) `(x)/(a)+(y)/(b)=1`,Parameter =2 .....(ii) `y-y_1=m(x-x_1)`, Parameter =2 ......(iii) and `xcosw+ysinw=p`, Parameter =2 .......(iv) it is clear that from Eqs. (i), (ii), (iii), and (iv), for specifying a straight line clearly two parameters should be known.

Discussion

136.	If the image of point `P(2, 3)` in a line `L` is `Q (4,5)` then, the image of point `R (0,0)` in the same line is:A. (3,2)B. (-2,3)C. (-3,-2)D. (3,-2)
Answer» Correct Answer - D The image of P(2,3) in the line mirror y=x is Q . So , the coordinates of Q are (3,2) . The image of Q(3,2) in the line mirror y =0 i.e. x-axis is (3,-2) . So , the coordinates of R are (3,-2).

Discussion

137.	If the lines `ax + by + c = 0, bx + cy + a = 0` and `cx + ay + b = 0` be concurrent, then:A. a + b = cB. b + c = aC. c + a = bD. a + b + c = 0
Answer» Correct Answer - D

Discussion

138.	The lines `p(p^2+1)x-y+q=0` and `(p^2+1)^2x+(p^2+1)y+2q=0` are perpendicular to a common line forA. no value of pB. exactly one value of pC. exactly two values of pD. more than two value of p
Answer» Correct Answer - B Given lines are perpendicular to a common line. So, they are parallel. Therefore, their slopw are equal. i.e., `p(p^(2)+1) = - (p^(2) + 1)rArr p = -1 " "[because p^(2) + 1 ne 0]` Hence, given lines are perpendicular to a common line for excatly one value of p.

Discussion

139.	The equation to a pair of opposite sides of a parallelogram are `x^2-5x+6=0`and `y^2+5=0`. The equations to its diagonalsare`x+4y=13 ,y=4x-7`(b) `4x+y=13 ,4y=x-7``4x+y=13 ,y=4x-7`(d)`y-4x=13 ,y+4x-7`A. `x+4y=13 and y=4x-7`B. `4x+y=13 and 4y=x-7`C. `4x+y=13 and y=4x-7`D. `y-4x=13 and y+4x=7`
Answer» Correct Answer - C

Discussion

140.	The equations of the sided of a triangle are `x+y-5=0,x-y+1=0,`and `x+y-sqrt(2)=0`is`(-oo,-4/3)uu(4/3,+oo)``(-4/3,4/3)`(c) `(-3/4,4/3)`none of theseA. `(-oo, -4//3) uu (4//3,oo)`B. `(-4//3,4//3)`C. `(-3//4,3//4)`D. none of these
Answer» Correct Answer - A

Discussion

141.	Find the angle which the line joining the points `(1, sqrt3) and (sqrt2, sqrt6)` makes with the x-axis.
Answer» Correct Answer - `60^(@)` `m=tan theta =((sqrt6-sqrt3)/(sqrt2-1))=sqrt3 Leftrightarrow theta=60^(@)`

Discussion

142.	A straight line L through the point (3,-2) is inclined at an angle `60^@` to the line `sqrt(3)x+y=1` If L also intersects the x-axis then the equation of L isA. `sqrt3 x + y + 2 - 3sqrt3 = 0`B. `y - sqrt3 x + 2 + 3sqrt3 = 0`C. `sqrt3 y - x + 3 + 2sqrt3 = 0`D. `sqrt3 y + x - 3 +2sqrt3 = 0`
Answer» The equations of lines passing through (3,-2) and inclined at `60^(@)` to the line `sqrt3 + y = 1` are given by `y + 2 = (-3 pm tan 60^(@))/(1 pm (-sqrt3) tan 60^(@)) (x - 3)` [Using : `y - y_(1) = (m pm tan 60^(@))/(1 pm m tan alpha ) ( x - x_(1))`] `implies y + 2 = (-sqrt3 pm sqrt3)/(1 pm 3) (x - 3)` `implies y + 2 = 0` and `y + 2 = sqrt3 (x - 3)` `implies y + 2 = 0` and `y - sqrt3 x + 2 + 3 sqrt3 = 0` Clearly `y - sqrt3 x + 2 + 3sqrt3 = 0` is the required line as it is not parallel to x-axis .

Discussion

143.	The equation of the straight line passing through the point `(4. 3)` and making intercepts on the co ordinate axes whose sum is ` -1`, isA. `(x)/(2) - (y)/(3) = 1 ` and `(x)/(-2) + (y)/(1) = 1`B. `(x)/(2) - (y)/(3) = -1` and `(x)/(-2) + (y)/(1) = -1`C. `(x)/(2) + (y)/(3) = 1` and `(x)/(2) + (y)/(1) = 1`D. `(x)/(2) + (y)/(3) = -1` and `(x)/(-2) + (y)/(1) = -1`
Answer» Correct Answer - A Let the equation of the line be `(x)/(a) + (y)/(b)=1 " " … (i)` It passses through (4,3) `therefore (4)/(a) + (3)/(b) = 1 " " … (ii)` It is given that a + b = - 1 `" " … (iii)` Solving these equations , we get (a =- 2 , b = 1) or , (a = 2 , b = -3) Substituting the value of a and b in (i) , we get `(x)/(-2) + (y)/(1) = 1` and `(x)/(2) + (y)/(-3) = 1` ALTER Clearly , lines in option (a) pass through (4,3) and the sum of their intercepts on the coordinates axes is -1 . Hence , option (a) is correct .

Discussion

144.	The locus of the orthocentre of the triangle formed by the lines `(1+p) x-py + p(1 + p) = 0, (1 + q)x-qy + q(1 +q) = 0` and y = 0, where `p!=*q`, is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight lineA. a hyperbolaB. a parabolaC. an ellipseD. a straight line
Answer» Correct Answer - D The intersection point y=0 with the first line si B(-p, 0). The intersection point of y=0 with the second line is A(-q, 0). The intersection point of the two line is C(pq, (p+1)(q+1) The altitude from C to AB is x = pq. The altitude from B to AC is `y = -(q)/(1+q)(x+p)` Solving these two, we get x=pq and y=-pq. Therefore, the locus of the orthocenter is x+y=0.

Discussion

145.	Find the equation of a horizontal line passing through the point (4,-2).
Answer» Correct Answer - y+2=0

Discussion

146.	Find the equation of a vertical line passing through the point (-5,6).
Answer» Correct Answer - x+5=0

Discussion

147.	Find the equation of a line passing through the origin and making an angle of `120^(@)` with the positive direction of the x-axis.
Answer» Correct Answer - `sqrt3x+y=0`

Discussion

148.	Find the equation of a line (i) whose slope is 4 and which passes through the point (5, -7), (ii) whose slope is -3 and which passes through the point (-2, 3), (iii) which makes an angle of `((2pi)/(3))` with the positive direction of the x-axis and passes through the point (0, 2).
Answer» Correct Answer - (i)4x-y=27=0 (ii)3x+y+3=0 (iii)`sqrt3+y-2=0`

Discussion

149.	A line intersects the straight lines `5x-y-4=0`and `3x-4y-4=0`at `A`and `B`, respectively. If a point `P(1,5)`on the line `A B`is such that `A P: P B=2:1`(internally), find point `Adot`
Answer» Correct Answer - (75/17, 307/17) Any point A on the first line is (t, 5t-4). Any point B on the second line is (r, (3r-4)/4). Hence, `1=(2r+t)/(3) " and " 5=((3r-4)/(2)+5t-4)/(3)` or 2r+t = 3 and 3r+10t = 42 On solving, we get `t = (75)/(17)` Hence, A is (75/17,307/17).

Discussion

150.	Find the slope and the equation of the line passing through the points: (i)(3,-2) and (-5,-7) (ii)(-1,1) and (2,-4) (5,3) and (-5,-3) (iv) (a,b) and (-a,b)
Answer» Correct Answer - `(i)(5)/(8), 5x-8y-31=0 (ii)(-5)/(3), 5x+3y+2=0` (iii)`(3)/(5),3x-5y=0 (iv)0,y=b`

Discussion

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Find angles between the lines `sqrt(3)x+y=1`and `x+sqrt(3)y=1`.

Point R (h, k) divides a line segment between the axes m the ratio `1: 2`. Find equation of the line.

If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.

Find the distance of the point (2, 3) from the line y=4.

Reduce the equation `x+y-sqrt2=0` to the normal form `x cos alpha+y sin alpha=p`, and hence find the values of `alpha and p`.

Reduce the equation `sqrt3+y+2=0` to the normal form `x cos alpha+y sin alpha=p`, and hence find the value of `alpha and p`.

Reduce each of the followringequations to normal form: (i)x+y-2=0 (ii) `x+y+sqrt2=0` (iii)x+5=0 (iv)2y-3=0 (v)4x+3y-9=0

Find the distance of the point (-4,3) from the line 4(x+5)=3(y-6)​

The coordinates of the image of the origin O with respect to the line `x+y+1=0` areA. (-1/2,-1/2)B. (-2,-2)C. (1,1)D. (-1,-1)

Reduce the equation y+4=0 to the normal form `x cos alhpha+y sin alpha=p` and hence find the values of `alpha and p`.

If the points `(1, 2)` and `(3, 4)` were to be on the opposite side of the `3x - 5y + alpha = 0`, then:A. `7 lt a lt 11`B. a=7C. a=1D. `a lt 7 " or " a gt 11`

Find the distance of the point (-2,3) from the line 12x=5y+13.

Find the distance of the point (4,1) from the line `3x-4y+12=0`

Find the distance of the point (-1,1) and the given line is `12x-5y+82=0`

Find the length of the perpendicular from the point (a,b) to the line `(x)/(a)+(y)/(b)=1`

Find the length of the perpendicular from the origin to the line 4x+3y-2=0

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that `1/(p^2)=1/(a^2)+1/(b^2)`.

Find the coordinates of the midpoint of the ilne segment joining the points, `A(-2,-5) and B(3,-1)`

Find what the following equation become when theorigin is shifted to the point (1,1): `x y-x-y+1=0`

Find what the following equation become when theorigin is shifted to the point (1,1): `x^2-y^2-2x+2y=0`

Find what the following equation become when theorigin is shifted to the point (1,1): `x y-y^2-x+y=0`

The equation `x^2+xy-3x-y+2=0` beome when the origin is shifted to the point `(1,1)` is

A line is drawn through the point (1, 2) to meet the coordinate axesat P and Q such that it forms a triangle OPQ, where O is the origin. If thearea of the triangle OPQ is least, then the slope of the line PQ is(1) `-1/4`(2) `-4`(3) `-2`(4) `-1/2`

If the point `A` is symmetric to the point `B(4,-1)` with respect to the bisector of the first quadrant then `AB` isA. `3sqrt(2)`B. `5sqrt(2)`C. `7sqrt(2)`D. `9sqrt(2)`

Let the algebraic sum of the perpendicular distances from the points `(2,0),(0,2)a n d(1,1)`to a variable straight line be zero. Then the line pass through a fixedpoint whose coordinates are`(1,1)`b. `(2,2)`c. `(3,3)`d. `(4,4)`A. `(-1,1)`B. `(1,1)`C. `(1,-1)`D. `(-1,-1)`

Find the equations of the lines, which cut-off intercepts on the axeswhose sum and product are `1`and `-6`, respectively.

`A(1,3)and c(-2/5,-2/5)`are the vertices of a `DeltaABCand`the equation of the angle bisector of `/_ABC` is `x+y=2.`A. (3/10, 17/10)B. (17/10, 3/10)C. (-5/2, 9/2)D. (1,1)

The point `(4,1)` undergoes the following two successive transformations (i) Reflection about the line `y=x` (ii) Translation through a distance 2 units along the positive X-axis. Then the final coordinate of the point areA. `(4,3)`B. `(3,4)`C. `(1,4)`D. `((7)/(2),(7)/(2))`

Find the equation of the line passing through the point P(-3,5) and perpendicular to the line passing through the points A(2, 5) and B(-3, 6).

If a, c, b are in G.P then the line `ax + by + c= 0`A. has a fixed directionB. always passes through a fixed pointC. forms a triangle with the axes whose area is constantD. always cuts intercepts on the axes such that their sum is zero

Straight lines `(x)/(a)+(y)/(b)=1, (x)/(b)+(y)/(a)=1,(x)/(a)+(y)/(b)=2 and (x)/(b)+(y)/(a)=2` form a rhombus of area ( in square units)A. `(ab)/(|a^(2)-b^(2)|)`B. `(ab)/(a^(2)+b^(2))`C. `(a^(2)b^(2))/(a^(2)+b^(2))`D. `(a^(2)b^(2))/(|a^(2)-b^(2)|)`

Equation of the line passing through `(1,2)` and parallel to the line `y=3x-1` isA. `y+2=x+1`B. `y+2=3(x+1)`C. `y-2=3(x-1)`D. `y-2=x-1`

Find the image of the point `(4,-13)`in the line `5x+y+6=0.`A. (1,-14)B. (6,-15)C. (-1,-14)D. none of these

For specifying a straight line, how many geomatrical parameters should be known ?A. 1B. 2C. 4D. 3

If the image of point `P(2, 3)` in a line `L` is `Q (4,5)` then, the image of point `R (0,0)` in the same line is:A. (3,2)B. (-2,3)C. (-3,-2)D. (3,-2)

If the lines `ax + by + c = 0, bx + cy + a = 0` and `cx + ay + b = 0` be concurrent, then:A. a + b = cB. b + c = aC. c + a = bD. a + b + c = 0

The lines `p(p^2+1)x-y+q=0` and `(p^2+1)^2x+(p^2+1)y+2q=0` are perpendicular to a common line forA. no value of pB. exactly one value of pC. exactly two values of pD. more than two value of p

The equation to a pair of opposite sides of a parallelogram are `x^2-5x+6=0`and `y^2+5=0`. The equations to its diagonalsare`x+4y=13 ,y=4x-7`(b) `4x+y=13 ,4y=x-7``4x+y=13 ,y=4x-7`(d)`y-4x=13 ,y+4x-7`A. `x+4y=13 and y=4x-7`B. `4x+y=13 and 4y=x-7`C. `4x+y=13 and y=4x-7`D. `y-4x=13 and y+4x=7`

The equations of the sided of a triangle are `x+y-5=0,x-y+1=0,`and `x+y-sqrt(2)=0`is`(-oo,-4/3)uu(4/3,+oo)``(-4/3,4/3)`(c) `(-3/4,4/3)`none of theseA. `(-oo, -4//3) uu (4//3,oo)`B. `(-4//3,4//3)`C. `(-3//4,3//4)`D. none of these

Find the angle which the line joining the points `(1, sqrt3) and (sqrt2, sqrt6)` makes with the x-axis.

A straight line L through the point (3,-2) is inclined at an angle `60^@` to the line `sqrt(3)x+y=1` If L also intersects the x-axis then the equation of L isA. `sqrt3 x + y + 2 - 3sqrt3 = 0`B. `y - sqrt3 x + 2 + 3sqrt3 = 0`C. `sqrt3 y - x + 3 + 2sqrt3 = 0`D. `sqrt3 y + x - 3 +2sqrt3 = 0`

The locus of the orthocentre of the triangle formed by the lines `(1+p) x-py + p(1 + p) = 0, (1 + q)x-qy + q(1 +q) = 0` and y = 0, where `p!=*q`, is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight lineA. a hyperbolaB. a parabolaC. an ellipseD. a straight line

Find the equation of a horizontal line passing through the point (4,-2).

Find the equation of a vertical line passing through the point (-5,6).

Find the equation of a line passing through the origin and making an angle of `120^(@)` with the positive direction of the x-axis.

Find the equation of a line (i) whose slope is 4 and which passes through the point (5, -7), (ii) whose slope is -3 and which passes through the point (-2, 3), (iii) which makes an angle of `((2pi)/(3))` with the positive direction of the x-axis and passes through the point (0, 2).

A line intersects the straight lines `5x-y-4=0`and `3x-4y-4=0`at `A`and `B`, respectively. If a point `P(1,5)`on the line `A B`is such that `A P: P B=2:1`(internally), find point `Adot`

Find the slope and the equation of the line passing through the points: (i)(3,-2) and (-5,-7) (ii)(-1,1) and (2,-4) (5,3) and (-5,-3) (iv) (a,b) and (-a,b)

Find the distance of the point (-4,3) from the line 4(x+5)=3(y-6)